Stress/Strain Lecture 1 ME EN 372 Andrew Ning aning@byu.edu Outline Stress Strain Plane Stress and Plane Strain Materials
otes and News [I had leftover time and so was also able to go through Section 3.1 at the end. Thi actually worked out well because the material is so related. I should probably plan on always going to that point.] Yesterday, the question came up: How do we know if we got the right answer? (in the context of the truss problem westress did for lab 1). Ask students what they think. In your careers, often you won t know for sure, but that doesn t mean we can t do any validation. Back-of-the-envelope calculations can be very useful. For this problem some suggestions include checking the reaction loads. Check the magnitude of th stresses, by doing a smaller problem. Assume all of the force is acting directly on tha member alone. Clearly the stress cannot be larger than that. In next week s lab we will discuss some debugging techniques in ANSYS and how to execute commands without the GUI..1 Stress 1D Stress Stress is a force per unit area. The SI units of stress are N/m 2 or Pa. In Imperia its stress is often given as pounds per square inch (psi). Figure 2.1 shows an example o e-dimensional stress in a bar given by the following formula = P A P P Figure 2.1: One-dimensional stress in a bar. More generally, stress is defined by both a direction of the stress and an orientation of th e upon which the stress acts. At a given point, stress is a tensor with nine component shown in Fig. 2.2 and represented mathematically as σ = P A 6
3D Stress State 2 3 = 4 xx xy xz yx yy yz 5 zx zy zz yy σ = σ xx τ xy τ xz τ yx σ yy τ yz τ zx τ zy σ zz zz zy yz zx yx xz xy xx Figure 2.2: The stress at any point in some material has nine components but only six independent (a symmetric second order tensor). The first index indicates the orientation of the face (e.g., x indicates a face whose norm points in the positive x direction), and the second index indicates the direction of the str Normal stress is denoted by and acts normal to a face. Normal stress involves tension compression. Sometimes, for convenience, the second index on normal stresses is dropp (i.e., x, y, z). Shear stress is denoted by and acts in the plane of the face. Shear str rotates or shears the structure. Figure 2.3 shows examples of pure tension, compression, a shear (rotation, while applicable to fluid dynamics, is generally irrelevant for structures) tension compression shear Figure 2.3 The stress tensor actually has only six independent components. From conservation angular momentum one can show that xz = zx, xy = yx,and yz = zy. Also, from sta equilibrium one can show that the stresses are equal and opposite on the opposite face. T positive sign conventions (for a 2D section) are shown in Fig. 2.4. 7
Positive Sign Convention y xy (+) x Figure 2.4: Positive sign convention for stresses. talk of nominal stress. This is the stress computed using the undeformed owever, you should keep in mind that there is an alternative: true stress he force and area are measured in the deformed configuration. For most lems the di erence Nominal is irrelevant. vs true Significant stress changes in area generally only ilure, the reduction in area is called necking. rmal and deformed] n of the stress tensor was given for a particular coordinate system, but you esulting stress in any direction. The force per unit area (often called the on a given surface with normal n is given by t = ˆn (2.1) y
Strain 1D Strain l 0 l ɛ = l l 0
u 0 u 1 x 0 x 1 ɛ x = u 1 u 0 u x 1 x 0 x Normal strain can occur in all three directions. ɛ x = u x ɛ y = v y ɛ z = w z
Analogous to shear stress, there are three independent shear strain components. γ xy = θ 1 + θ 2 2 1 γ xy = γ yx = v x + u y γ yz = γ zy = w y + v z γ xz = γ zx = u z + w x
Plane Stress and Plane Strain Plane Stress 2 3 = 4 xx xy xz yx yy yz 5 zx zy zz yy y yz yx xy xy zy xx (+) x zx xz zz Figure 2.4: Positive sign convention for stresses. igure 2.2: The stress at any point in some material has nine components but only six are dependent (a symmetric second order tensor). Generally we talk of nominal stress. This is the stress computed using th configurations. However, you should keep in mind that there is an alternati The first index indicates the orientation inofwhich the face both (e.g., the x indicates force and a face area whose arenormal measured in the deformed configurat oints in the positive x direction), and the engineering second index problems indicates the the direction di erence of the is irrelevant. stress. Significant changes in area
Plane Strain Materials
Homogeneous: Isotropic: Linear Stress-Strain Relationships Most general material (in the elastic region): ɛ xx ɛ yy ɛ zz γ xy γ xz γ yz = S 11 S 12 S 13 S 14 S 15 S 16 S 21 S 22 S 23 S 24 S 25 S 26 S 31 S 32 S 33 S 34 S 35 S 36 S 41 S 42 S 43 S 44 S 45 S 46 S 51 S 52 S 53 S 54 S 55 S 56 S 61 S 62 S 63 S 64 S 65 S 66 σ xx σ yy σ zz τ xy τ xz τ yz
ɛ = Sσ S: compliance matrix Inverse: σ = Kɛ K: stiffness matrix. Orthotropic Materials Usually, a material contain certain symmetries. ɛ xx ɛ yy ɛ zz γ xy γ xz γ yz = S 11 S 12 S 13 0 0 0 S 21 S 22 S 23 0 0 0 S 31 S 32 S 33 0 0 0 0 0 0 S 44 0 0 0 0 0 0 S 55 0 0 0 0 0 0 S 66 σ xx σ yy σ zz τ xy τ xz τ yz
Isotropic A special case of orthotropy is isotropy, in which the elastic properties are the same in every direction. ɛ x ɛ y ɛ z γ xy γ xz γ yz = 1 E 1 ν ν 0 0 0 ν 1 ν 0 0 0 ν ν 1 0 0 0 0 0 0 2(1 + ν) 0 0 0 0 0 0 2(1 + ν) 0 0 0 0 0 0 2(1 + ν) σ x σ y σ z τ xy τ xz τ yz Inverse (stiffness matrix) σ x σ y σ z τ xy τ xz τ yz = E (1 + ν)(1 2ν) 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 ν ν 1 ν 0 0 0 1 2ν 0 0 0 2 0 0 1 2ν 0 0 0 0 2 0 0 0 0 0 0 1 2ν 2 ɛ x ɛ y ɛ z γ xy γ xz γ yz
G = E 2(1 + ν) Strain in x-direction ɛ x = 1 E [σ x ν(σ y + σ z )]