ASEN 3112 - Structures Stress in 3D ASEN 3112 Lecture 1 Slide 1
ASEN 3112 - Structures Mechanical Stress in 3D: Concept Mechanical stress measures intensit level of internal forces in a material (solid or fluid bod) idealied as a mathematical continuum. The phsical measure of stress is Force per unit area 2 e.g. N/mm (MPa) or lbs/sq-in (psi) This measure is convenient to assess the resistance of a material to permanent deformation (ield, creep, slip) and rupture (fracture, cracking). Comparing working and failure stress levels allows engineers to establish strength safet factors for structures. Stresses ma var from point to point. We net consider a solid bod (could be a structure or part of one) in 3D. ASEN 3112 Lecture 1 Slide 2
Cutting a 3D Bod ASEN 3112 - Structures Consider a 3D solid bod in static equilibrium under applied loads. We want to find the state of stress at an arbitrar point Q, which generall will be inside the bod. Cut the bod b a plane ABCD that passes through Q as shown (How to orient the plane is discussed later.) The bod is divided into two. Retain one portion (red in figure) and discard the other (blue in figure) To restore equilibrium, however, we must replace the discarded portion b the internal forces it had eerted on the kept portion. Applied loads cut b plane ABCD & discard blue portion Applied loads Keep A D ;Q Cut plane Q B C Applied loads Discard ASEN 3112 Lecture 1 Slide 3
Orienting the Cut Plane The cut plane ABCD is oriented b its unit normal direction vector n, or normal for short. B convention we will draw n as emerging from Q and pointing outward from the kept bod. This direction identifies the eterior normal. With respect to the RCC sstem {,,}, the normal vector has components n = n n n where {n,n,n } are the direction cosines of n with 2 2 2 respect to {,,}. These satisf n + n + n = 1. In the figure, the cut plane ABCD has been chosen with its eterior normal parallel to the + ais. Consequentl Applied loads n = A Kept bod ;; ; Q D Reference frame is a right-handed, RCC sstem 1 0 0 Cut plane ASEN 3112 - Structures B n // C Unit vector n is the eterior normal at Q, meaning it points outward from the kept bod ASEN 3112 Lecture 1 Slide 4
ASEN 3112 - Structures Digression: Action and Reaction (Newton's 3rd Law, from Phsics I) Remove the "bridge" (log) and replace its effect on the elephant b reaction forces on the legs. The elephant stas happ: nothing happens. ;; ;;; ;;; Strictl speaking, reaction forces are distributed over the elephant leg contact areas. The are replaced above b equivalent point forces, a.k.a. resultants, for visualiation convenience ASEN 3112 Lecture 1 Slide 5
Internal Forces on Cut Plane ASEN 3112 - Structures Those internal forces generall will form a sstem of distributed forces per unit of area, which, being vectors, generall will var in magnitude and direction as we move from point to point of the cut plane, as pictured. Applied loads Q ; Internal forces Net, we focus our attention on point Q. Pick an elemental area A around Q that lies on the cut plane. Call the resultant of the internal forces that act on A. Draw that vector with origin at Q, as pictured. Don't forget the normal The use of the increment smbol suggests a pass to the limit. This will be done later to define the stresses at Q Applied loads ;; Q A Arrows are placed over and n to remind ou that the are vectors n ASEN 3112 Lecture 1 Slide 6
Internal Force Components Zoom on the elemental area about Q, omitting the kept-bod and applied loads for clarit: ASEN 3112 - Structures Project vector on aes, and to get its components, and, respectivel. See bottom figure. Q A n Note that component is aligned with the cut-plane normal, because n is parallel to. It is called the normal internal force component. Components and lie on the cut plane. The are called tangential internal force components. A Q We are now read to define stresses. n // ASEN 3112 Lecture 1 Slide 7
ASEN 3112 - Structures Stress Components at a Point: Cut Define the -stress components at point Q b taking the limits of internal-force-over-area ratios as the elemental area shrinks to ero: Q A (Reproduced from previous slide for convenience) n // F F σ def lim def = = lim def = A 0 A A 0 A F lim A 0 A σ is called a normal stress, whereas and are shear stresses. ASEN 3112 Lecture 1 Slide 8
Stress Components at a Point: and Cuts It turns out we need nine stress components in 3D to full characterie the stress state at a point. So far we got onl three. Si more are obtained b repeating the same take-the-limit procedure with two other cut planes. The obvious choice is to pick planes normal to the other two aes: and. Taking n// we get three more components, one normal and two shear: ASEN 3112 - Structures σ These are called -stress components. Taking n// we get three more components, one normal and two shear: σ These are called -stress components. Together with the three -stress components found before, this makes up a total of nine, as required. ASEN 3112 Lecture 1 Slide 9
ASEN 3112 - Structures Visualiation on "Stress Cube" The foregoing nine stress components ma be convenientl visualied on a "stress cube" as follows. Cut an infinitesimal cube d d d around Q with sides parallel to the RCC aes {,,}. Draw the components on the positive cube faces (defined below) as σ Point Q is inside cube + face d + face + face σ σ d d Note that stresses are forces per unit area, not forces, although the look like forces in the picture. Strictl speaking, this is a "cube" onl if d=d=d, else it should be called a parallelepided; but that is difficult to pronounce. The three positive cube faces are those with eterior (outward) normals aligned with +, + and +, respectivel. Positive (+) values for stress components on those faces are as shown. More on sign conventions later. ASEN 3112 Lecture 1 Slide 10
What Happens On The Negative Faces? ASEN 3112 - Structures The stress cube has three positive (+) faces. The three opposite ones are negative ( ) faces. Outward normals at faces point along, and. What do stresses on those faces look like? To maintain static equilibrium, stress components must be reversed. For eample, a positive σ points along + on the + face, but along on the face. A positive points along + on the + face but along on the face. To visualie the reversal, it is convenient to project the stress cube onto the {,} plane b looking at it from the + direction. The resulting 2D figure clearl displas the "reversal recipe" given above. σ σ σ + face face σ project on {,} σ σ + face σ face ASEN 3112 Lecture 1 Slide 11
ASEN 3112 - Structures Notational & Sign Conventions Shear stress components have two indices. The first one identifies the cut plane on which it acts, through the normal to that plane. The second inde identifies component direction. For eample: stress acts on cut plane with n along + (the + face) Stress sign conventions are as follows. stress points in the direction For a normal stress: positive (negative) if it produces tension (compression) in the material. + face For a shear stress: positive if, when acting on the + face identified b the first inde, it points in the + direction identified b the second inde. Eample: is + if on the + face it points in the + direction; see figure. The sign of a shear stress has no phsical meaning; it is entirel conventional. σ Both σ and are + as drawn above ASEN 3112 Lecture 1 Slide 12
Matri Representation of Stress ASEN 3112 - Structures σ The nine components of stress referred to the {,,} aes ma be arranged as a 3 3 matri, which is configured as σ σ σ σ σ Note that normal stresses are placed in the diagonal of this square matri. We will call this a 3D stress matri, although in more advanced courses this is the representation of a second-order tensor. ASEN 3112 Lecture 1 Slide 13
Shear Stress Reciprocit ASEN 3112 - Structures σ From moment equilibrium conditions on the stress cube it can be shown that =, =, = σ σ in magnitude. In other words: switching shear stress indices does not change its value. Note, however, that stresses point in different directions:, sa, points along whereas points along. This propert is known as shear stress reciprocit. It follows that the stress matri is smmetric: σ = = σ = σ σ σ σ smm Consequentl the 3D stress state depends on onl si (6) independent parameters: three normal stresses and three shear stresses. ASEN 3112 Lecture 1 Slide 14
Simplifications: 2D and 1D Stress States ASEN 3112 - Structures For certain structural configurations such as thin plates, all stress components with a subscript ma be considered negligible, and set to ero. The stress matri becomes σ σ 0 0 0 0 0 This 2D simplification is called plane stress state. Since =, this state is characteried b three independent stress components: σ, σ and =. A further simplification occurs in structures such as bars or beams, in which all stress components ecept σ ma be considered negligible and set to ero, whence the stress matri reduces to σ 0 0 0 0 0 0 0 0 This is called a one-dimensional stress state. There is onl one independent stress component left: σ ASEN 3112 Lecture 1 Slide 15
ASEN 3112 - Structures Changing Coordinate Aes (1) Suppose we change aes {,,} to another set {',','} that also forms a RCC sstem. The stress cube centered at Q is rotated to realign with {',','} as pictured below. ' ' ' σ d d σ The stress components change accordingl, as shown in matri form: d σ Note that Q is inside both cubes; the are drawn offset for clarit becomes σ' ' ' d' σ' ' ' d' ' ' d' ' ' σ' ' Can the primed components be epressed in terms of the original ones? ASEN 3112 Lecture 1 Slide 16
ASEN 3112 - Structures Changing Coordinate Aes (2) The answer is es. All primed stress components can be epressed in terms of the unprimed ones and of the direction cosines of {',','} with respect to {,,}. This operation is called a stress transformation. For a general 3D state this operation is quite complicated because there are three direction cosines. In this introductor course we will cover onl transformations for the 2D plane stress state. These are simpler because changing aes in 2D depends on onl one direction cosine or, equivalentl, the rotation angle about the ais. Wh do we bother to look at stress transformations? Well, material failure ma depend on the maimum normal tensile stress (for brittle materials) or the maimum absolute shear stress (for ductile materials). To find those we generall have to look at parametric rotations of the coordinate sstem, as in the skew-cut bar eample studied later. Once such dangerous stress maima are found for a given structure, the engineer can determine strength safet factors. ASEN 3112 Lecture 1 Slide 17
ASEN 3112 Lecture 1 Slide 18