Plane Stress Transformations

Transcription

1 6 Plane Stress Transformations 6 1

2 Lecture 6: PLANE STRESS TRANSFORMATIONS TABLE OF CONTENTS Page 6.1 Introduction Thin Plate in Plate Stress D Stress Transformations Wh Are Stress Transformations Important? Method of Equations Double Angle Version Principal Stresses, Planes, Directions, Angles Maimum Shear Stresses Principal Stress Element Mohr s Circle What Happens in 3D? Including the Plane Stress Thickness Dimension D Mohr Circles Overall Maimum Shear Plane Stress Revisited The Sphere Parado

3 6. THIN PLATE IN PLATE STRESS 6.1. Introduction This Lecture deals with the plate problem. This is a two-dimensional state, briefl introduced in? of Lecture 1. It occurs frequentl in two kinds of aerospace structural components: 1. Thin wall plates and shells; e.g., aircraft and rocket skins, and the pressure vessels of Lecture 3.. Shaft members that transmit torque. These will be studied in Lectures 7 9. The material below focuses on thin flat plates, and works out the associated problem of plane transformations. 6.. Thin Plate in Plate Stress In structural mechanics, a flat thin sheet of material is called a plate. The distance between the plate faces is the thickness, which is denoted b h. The midplane lies halfwa between the two faces. The direction normal to the midplane is the transverse z direction. Directions parallel to the midplane are called Top surface in-plane directions. The global ais z is oriented along the transverse direction. Aes and are placed in the midplane, forming a right-handed Rectangular Cartesian Coordinate (RCC) sstem. Thus the equation of the midplane is z =. The +z ais conventionall defines the top surface of the plate as the one that it intersects, whereas the opposite surface is called the Figure 6.1. A plate structure in plane. bottom surface. See Figure 6.1. A plate loaded in its midplane is said to be in a state of plane, oramembrane state, if the following assumptions hold: 1. All loads applied to the plate act in the midplane direction, as pictured in Figure 6.1, and are smmetric with respect to the midplane.. All support conditions are smmetric about the midplane. 3. In-plane displacements, strains and es are taken to be uniform through the thickness. 4. The normal and shear components in the z direction are zero or negligible. The last two assumptions are not necessaril consequences of the first two. For those to hold, the thickness h should be small, tpicall 1% or less, than the shortest in-plane dimension. If the plate thickness varies it should do so graduall. Finall, the plate fabrication must ehibit smmetr with respect to the midplane. To these four assumptions we add an adscititious restriction: 5. The plate is fabricated of the same material through the thickness. Such plates are called transversel homogeneous or (in aerospace) monocoque plates. The last assumption ecludes wall constructions of importance in aerospace, in particular composite and honecomb sandwich plates. The development of mathematical models for such configurations requires a more complicated integration over the thickness as well as the abilit to handle coupled bending and stretching effects. Those topics fall outside the scope of the course. 6 3

4 Lecture 6: PLANE STRESS TRANSFORMATIONS Midplane Mathematical idealization Plate Γ Ω Figure 6.. Mathematical model of plate in plane. (Smbols and Ɣ, used to denote the plate interior and the boundar, respectivel, are used in advanced courses.) Thin plate in plane z In-plane internal forces h d d p p p d d + sign conventions for internal forces, es and strains d d h In-plane es σ d d τ = τ σ In-plane bod forces h b d d b In-plane strains d d h ε ε γ = γ In-plane displacements h u d d u Figure 6.3. Notational conventions for in-plane es, strains, displacements and internal forces of a thin plate in plane. Remark 6.1. Selective relaation from assumption 4 leads to the so-called generalized plane state, in which nonzero σ zz es are accepted; but these es do not var with z. The plane strain state described in? of Lecture 5 is obtained if strains in the z direction are precluded: ɛ zz = γ z = γ z =. Remark 6.. Transverse loading on a plate produces plate bending, which is associated with a more comple configuration of internal forces and deformations. This topic is studied in graduate-level courses. Remark 6.3. The requirement of flatness can be relaed to allow for a curved configuration, as long as the structure, or structure component, resists primaril in-plane loads. In that case the midplane becomes a midsurface. Eamples are rocket and aircraft skins, ship and submarine hulls, open parachutes, boat sails and balloon walls. Such configurations are said to be in a membrane state. Another eample are thin-wall members under torsion, which are covered in Lectures 8 9. The plate in plane idealized as a two-dimensional problem is illustrated in Figure 6.. In this idealization the third dimension is represented as functions of and that are integrated 6 4

5 6.3 D STRESS TRANSFORMATIONS (a) P σ τ τ σ (b) σ tt τ tn P τ nt σ nn Global aes, sta fied z t z θ n Local aes n,t rotate b θ Figure 6.4. Plane sstem referred to global aes, (data) and to local rotated aes n, t. (Locations of point P in (a,b) coincide; the are drawn offset for visualization convenience.) through the plate thickness. Engineers often work with internal plate forces, which result from integrating the in-plane es through the thickness. See Figure 6.3. In this Lecture we focus on the in-plane es σ, σ and τ and their epressions with respect to an arbitrar sstem of aes 6.3. D Stress Transformations The transformation problem studied in this Lecture is illustrated in Figure 6.4. Stress components σ, σ and τ at a midplane point P are given with respect to the global aes and, as shown in Figure 6.4(a). The material element about P is rotated b an angle θ that aligns it with aes n, t, as shown in Figure 6.4(b). (Note that location of point P in (a,b) coincide the are drawn offset for visualization convenience.) The transformation problem consists of epressing σ nn, σ tt, and τ nt = τ tn in terms of the data σ, σ and τ, and of the angle θ. Two methods, one analtical and one graphical, will be described here. Before that is done, it is useful to motivate what are the main uses of these transformations Wh Are Stress Transformations Important? The transformation problem has two major uses in structural analsis and design. Find es along a given skew direction. Here θ is given as data. This has several applications. Two eamples: 1. Analsis of fiber reinforced composites if the direction of the fibers is not aligned with the {, } aes. A tensile normal perpendicular to the fibers ma cause delamination, and a compressive one ma trigger local buckling.. Oblique joints that ma fail b shear parallel to the joint. For eample, welded joints. Find ma/min normal es, ma in-plane shear and overall ma shear. This ma be important for strength and safet assessment. Here finding the angle θ is part of the problem. Both cases are covered in the following subsections. 6 5

6 Lecture 6: PLANE STRESS TRANSFORMATIONS Method of Equations The derivation given on pages of Vable or in 7. of Beer-Johnston-DeWolf is based on the wedge method. This will be omitted here for brevit. The final result is σ nn = σ cos θ + σ sin θ + τ sin θ cos θ, σ tt = σ sin θ + σ cos θ τ sin θ cos θ, τ nt = (σ σ ) sin θ cos θ + τ (cos θ sin θ). (6.1) Couple of checks are useful to verif these equations. If θ =, σ nn = σ, σ tt = σ, τ nt = τ, (6.) as epected. If θ = 9, σ nn = σ, σ tt = σ, τ nt = τ. (6.3) which is also OK (can ou guess wh τ nt at θ = 9 is τ?). Note that σ + σ = σ nn + σ tt. (6.4) This sum is independent of θ, and is called a invariant. (Mathematicall, it is the trace of the tensor.) Consequentl, if σ nn is computed, the fastest wa to get σ tt is as σ + σ σ nn, which does not require trig functions Double Angle Version For man developments it is convenient to epress the transformation equations (6.1) in terms of the double angle θ b using the well known trigonometric relations cos θ = cos θ sin θ and sin θ = sin θ cos θ, in addition to sin θ + cos θ = 1. The result is σ nn = σ + σ τ nt = σ σ + σ σ sin θ + τ cos θ. cos θ + τ sin θ, (6.5) Here σ tt is omitted since, as previousl noted, it can be quickl computed as σ tt = σ + σ σ nn Principal Stresses, Planes, Directions, Angles The maimum and minimum values attained b the normal σ nn, considered as a function of θ, are called principal es. (A more precise name is in-plane principal normal es, but the qualifiers in-plane and normal are often dropped for brevit.) Those values occur if the derivative dσ nn /dθ vanishes. Differentiating the first of (6.1) and passing to double angles gives dσ nn dθ = (σ σ ) sin θ cos θ + τ (cos θ sin θ) = (σ σ ) sin θ + τ cos θ =. 6 6 (6.6)

7 6.3 D STRESS TRANSFORMATIONS This is satisfied for θ = θ p if tan θ p = τ σ σ (6.7) There are two double-angle solutions θ 1 and θ given b (6.7) in the range [ θ 36 ]or [ 18 θ 18 ] that are 18 apart. (Which θ range is used depends on the tetbook; here we use the first one.) Upon dividing those values b, the principal angles θ 1 and θ define the principal planes, which are 9 apart. The normals to the principal planes define the principal directions. Since the differ b a 9 angle of rotation about z, it follows that the principal directions are orthogonal. As previousl noted, the normal es that act on the principal planes are called the in-plane principal normal es, or simpl principal es. The are denoted b σ 1 and σ, respectivel. Using (6.7) and trigonometric relations it can be shown that their values are given b σ 1, = σ + σ ± (σ σ ) + τ. (6.8) To evaluate (6.8) it is convenient to go through the following staged sequence: 1. Compute σ av = σ (σ ) + σ σ and R =+ + τ. (6.9) Meaning of these values: σ av is the average normal at P (recall that σ + σ is an invariant and so is σ av ), whereas R is the radius of the Mohr s circle, as described in below, thus the smbol. Furthermore, R represents the maimum in-plane shear value, as discussed in below.. The principal values are σ 1 = σ av + R, σ = σ av R. (6.1) 3. Note that the a priori computation of the principal angles is not needed to get the principal es if one follows the foregoing steps. If finding those angles is of interest, use (6.7). Comparing (6.6) with the second of (6.5) shows that dσ nn /dθ = τ nt. Since dσ nn /dθ vanishes for a principal angle, so does τ nt. Hence the principal planes are shear free Maimum Shear Stresses Planes on which the maimum shear es act can be found b setting dτ nt /dθ =. A stud of this equation shows that the maimum shear planes are located at ±45 from the principal planes, and that the maimum and minimum values of τ nt are ±R. See for eample, 7.3 of the Beer-Johnston-DeWolf tetbook. This result can be obtained graphicall on inspection of Mohr s circle, covered later. 6 7

8 Lecture 6: PLANE STRESS TRANSFORMATIONS (a) P σ = psi τ = τ =3 psi σ =1 psi t σ =1 psi (b) n θ P principal directions θ = σ =11 psi 1 θ 1=18.44 principal planes (d) plane of ma inplane shear (c) 45 τ = R=5 psi ma P principal element P = principal planes principal planes Figure 6.5. Plane eample: (a) given data: components σ, σ and τ ; (b) principal es and angles; (c) maimum shear planes; (d) a principal element (actuall four PSE can be drawn, that shown is one of them). Note: locations of point P in (a) through (d) coincide; the are drawn offset for visualization convenience. Eample 6.1. This eample is pictured in Figure 6.5. Given: σ = 1 psi, σ = psi and τ = 3 psi, as shown in Figure 6.5(a), find the principal es and their directions. Following the recommended sequence (6.9) (6.1), we compute first ( ) σ av = = 6 psi, R =+ + 3 = 5 psi, (6.11) from which the principal es are obtained as σ 1 = = 11 psi, σ = 6 5 = 1 psi. (6.1) To find the angles formed b the principal directions, use (6.7): tan θ p = 3 1 = 3 4 =.75, with solutions θ 1 = 36.87, θ = θ = 16.87, (6.13) whence θ 1 = 18.44, θ = θ = These values are shown in Figure 6.5(b). As regards maimum shear es, we have τ ma =R = 5 psi. The planes on which these act are located at ±45 from the principal planes, as illustrated in Figure 6.5(c) Principal Stress Element Some authors, such as Vable, introduce here the so-called principal element or PSE. This is a wedge formed b the two principal planes and the plane of maimum in-plane shear. Its projection on the {, } plane is an isosceles triangle with one right angle and two 45 angles. For the foregoing eample, Figure 6.5(d) shows a PSE. There are actuall 4 was to draw a PSE, since one can join point P to the opposite corners of the square aligned with the principal planes in two was along diagonals, and each diagonal splits the square into two triangles. The 4 images ma be sequentiall produced b appling sucessive rotations of 9. Figure 6.5(d) shows one of the 4 possible PSEs for the eample displaed in that Figure. The PSE is primaril used for the visualization of material failure surfaces in fracture and ield, a topic onl covered superficiall here. 6 8

9 6.3 D STRESS TRANSFORMATIONS (a) Point in plane (a) P σ = psi τ = τ =3 psi σ =1 psi σ = τ = shear H C θ = = τ ma = 5 Radius R = 5 V σ = normal σ = 11 1 θ 1 = τ min = 5 (b) Mohr's circle coordinates of blue points are H: (,3), V:(1,-3), C:(6,) Figure 6.6. Mohr s circle for plane eample of Figure Mohr s Circle Mohr s circle is a graphical representation of the plane state at a point. Instead of using the methods of equations, a circle is drawn on the {σ, τ} plane. The normal σ(θ) and the share τ(θ) are plotted along the horizontal and vertical aes, respectivel, with θ as a parameter. All states obtained as the angle θ is varied fall on a circle called Mohr s circle. 1 This representation was more important for engineers before computers and calculators appeared. But it still retains some appealing features, notabl the clear visualization of principal es and maimum shear. It also remains important in theories of damage, fracture and plasticit that have a failure surface. To eplain the method we will construct the circle corresponding to Eample 6.1, which is reproduced in Figure 6.6(a) for convenience. Draw horizontal ais σ = σ nn (θ) to record normal es, and vertical ais τ = τ nt (θ) to record shear es. Mark two points: V (for vertical cut, meaning a plane with eterior normal parallel to ) at(σ, τ ) ={1, 3} and H (for horizontal cut, meaning a plane with eterior normal parallel to ) at(σ,τ ) = (, 3). The midpoint between H and V is C, the circle center, which is located at ( 1 (σ + σ ), ) = (σ av, ) = (6, ). Now draw the circle. It ma be verified that its radius is R as given b (6.9); which for Eample 6.1 is R = 5. See Figure 6.6(b). The circle intersects the σ ais at two points with normal es σ avg + R = 11 = σ 1 and σ av R = 1 = σ. Those are the principal es. Wh? At those two points the shear τ nt vanishes, which as we have seen characterizes the principal planes. The maimum in-plane shear occurs when τ = τ nt is maimum or minimum, which happens at the highest and lowest points of the circle. Obviousl τ ma = R = 5 and τ min = R = 5. What is the normal when the shear is maimum or minimum? B inspection it is σ av = 6 because those points lie on a vertical line that passes through the circle center C. 1 Introduced b Christian Otto Mohr (a civil engineer and professor at Dresden) in 188. Other important personal contributions were the concept of staticall indeterminate structures and theories of material failure b shear. 6 9

10 Lecture 6: PLANE STRESS TRANSFORMATIONS Other features, such as the correlation between the angle θ on the phsical plane and the rotation angle θ traversed around the circle will be eplained in class if there is time. If not, one can find those details in Chapter 7 of Beer-Johnston-DeWolf tetbook, which covers Mohr s circle well What Happens in 3D? Despite the common use of simplified 1D and D structural models, the world is three-dimensional. Stresses and strains actuall live in 3D. When the etra(s) space dimension(s) are accounted for, some paradoes are resolved. In this final section we take a quick look at principal es in 3D, stating the major properties as recipes Including the Plane Stress Thickness Dimension To fi the ideas, 3D results will be linked to the plane case studied in Eample 6.1. and pictured in Figure 6.5. Its 3D state of in {,, z} coordinates is defined b the 3 3 matri [ 1 3 ] S = 3 (6.14) It ma be verified that the eigenvalues of this matri, arranged in descending order, are σ 1 = 11, σ = 1, σ 3 =. (6.15) Now in 3D there are three principal es, which act on three mutuall orthogonal principal planes that are shear free. For the matri (6.14) the principal values are 11, 1 and. But those are precisel the eigenvalues in (6.15). This result is general: The principal es in 3D are the eigenvalues of the 3D matri The matri is smmetric. A linear algebra theorem sas that a real smmetric matri has a full set of eigenvalues and eigenvectors, and that both eigenvalues and eigenvectors are guaranteed to be real. The normals to the principal planes (the so-called principal directions) are defined b the three orthonormalized eigenvectors, but this topic will not be pursued further. In plane, one of the eigenvalues is alwas zero because the last row and column of S are null; thus one principal is zero. The associated principal plane is normal to the transverse ais z, as can be phsicall epected. Consequentl σ zz = is a principal. The other two principal es are the in-plane principal es, which were those studied in It ma be verified that their values are given b (6.8), and that their principal directions lie in the {, } plane. Some terminolog is needed. Suppose that the three principal es are ordered, as usuall done, b decreasing algebraic value: σ 1 σ σ 3 (6.16) Then σ 1 is called the maimum principal, σ 3 the minimum principal, and σ the intermediate principal. (Note that this ordering is b algebraic, rather than b absolute, value.) In the plane eample (6.14) the maimum, intermediate and minimum normal es are 11, 1 and, respectivel. The first two are the in-plane principal es. 6 1

11 6.4 WHAT HAPPENS IN 3D? (a) τ = shear σ 3 Inner Mohr's circles σ Overall ma shear All possible states lie in the shaded area between the outer and inner circles σ = normal σ 1 Outer Mohr's circle (b) σ = 3 σ = 1 τ = shear overall τ ma = 55 inplane τ = 5 ma σ = normal σ = 11 1 Yellow-filled circle is the inplane one Figure 6.7. Mohr s circles for a 3D state: (a) general case; (b) plane eample of Figure 6.5. In (b) the Mohr circle of Figure 6.6 is the rightmost inner circle. Actual states lie on the gre shaded areas D Mohr Circles More surprises (pun intended): in 3D there are actuall three Mohr circles, not just one. To draw the circles, start b getting the eigenvalues σ 1, σ and σ 3 of the matri, for eample through the eig function of Matlab. Suppose the are ordered as per the convention (6.16). Mark their values on the σ ais of the σ vs. τ plane. Draw the 3 circles with diameters {σ 1,σ }, {σ,σ 3 } and {σ 1,σ 3 }, as sketched in Figure 6.7(a). For the principal values (6.15) the three Mohr circles are drawn in Figure 6.7(b), using a scale similar to that of Figure 6.6. Of the three circles the wider one goes from σ 1 (maimum normal ) to σ 3 (minimum normal ). This is known as the outer or big circle, while the two others are called the inner circles. It can be shown (this is proven in advanced courses in continuum mechanics) that all actual states at the material point lie between the outer circle and the two inner ones. Those are marked as gre shaded areas in Figure Overall Maimum Shear For ductile materials such as metal allos, which ield under shear, the 3D Mohr-circles diagram is quite useful for visualizing the overall maimum shear at a point, and hence establish the factor of safet against that failure condition. Looking at the diagram of Figure 6.7(a), clearl the overall maimum shear, called τma overall, is given b the highest and lowest point of the outer Mohr s circle, marked as a blue dot in that figure. (Note that onl its absolute value is of importance for safet checks; the sign has no importance.) But that is also equal to the radius R outer of that circle. If the three principal es are algebraicall ordered as in (6.16), then τma overall = R outer = σ 3 σ 1 Note that the intermediate principal σ does not appear in this formula. (6.17) In the general 3D case there is no simple geometric construction of the circles starting from the si,, z independent components, as done in for plane. 6 11

12 Lecture 6: PLANE STRESS TRANSFORMATIONS If the principal es are not ordered, it is necessar to use the ma function in a more complicated formula that selects the largest of the 3 radii: ( ) τma overall σ 1 σ = ma, σ σ 3, σ 3 σ 1 (6.18) Taking absolute values in (6.18) is important because the ma function picks up the largest algebraic value. For eample, writing τma overall = ma(3, 5, ) = 3 picks up the wrong value. On the other hand τma overall ma( 3, 5, ) = 5 is correct Plane Stress Revisited Going back to plane, how do overall maimum shear and in-plane maimum shear compare? Recall that one of the principal es is zero. The ordering (6.16) of the principal es is assumed, and one of those three is zero. The following cases ma be considered: (A) The zero is the intermediate one: σ =. If so, the in-plane Mohr circle is the outer one and the two shear maima coincide: τ overall ma = τ inplane ma = σ 1 σ 3 (6.19) (B) The zero is either the largest one or the smallest one. Two subcases: (B1) If σ 1 σ and σ 3 = : (B) If σ 3 σ and σ 1 = : τ overall ma = σ 1, τ overall ma = σ 3. (6.) In the plane eample of Figure 6.5, the principal es are given b (6.1). Since both in-plane principal es (11 psi and 1 psi) are positive, the zero principal is the smallest one. We are in case (B), subcase (B1). Consequentl τma overall = 1 σ 1 = 55 psi, whereas τma inplane = 1 (σ 1 σ ) = 5 psi. 6 1

13 6.4 WHAT HAPPENS IN 3D? (a) τ = shear inplane τ ma = σ = σ =8 1 σ = normal (b) σ = 3 τ = shear overall τ ma = 4 inplane τ ma = σ = σ =8 1 σ = normal Figure 6.8. The sphere parado: (a) Mohr s in-plane circle reduced to a point, whence τma inplane = ; (b) drawing the 3D circles shows that τma overall = The Sphere Parado Taking the third dimension into account clarifies some puzzles such as the sphere parado. Consider a thin spherical pressure vessel with p = 16 psi, R = 1 in and t =.1 in. In Lecture 3, the wall in spherical coordinates was found to be σ = pr/( t) = 8 ksi, same in all directions. The matri at an point in the wall, taking z as the normal to the sphere at that point, is [ ] 8 8 (6.1) The Mohr circle reduces to a point, as illustrated in Figure 6.8(a), and the maimum in-plane shear is zero. And remains zero for an p. If the sphere is fabricated of a ductile material, it should never break regardless of pressure. Plainl we have a parado. The parado is resolved b considering the other two Mohr circles. These additional circles are pictured in Figure 6.8(b). In this case the outer circle and the other inner circle coalesce. The overall maimum shear is 4 ksi, which is nonzero. Therefore, increasing the pressure will eventuall produce ield, and the parado is resolved. 6 13