P4 Stress and Strain Dr. A.B. Zavatsky HT08 Lecture 5 Plane Stress Transformation Equations Stress elements and plane stress. Stresses on inclined sections. Transformation equations. Principal stresses, angles, and planes. Maximum shear stress. 1 Normal and shear stresses on inclined sections To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an “inclined” (as opposed to a “normal”) section through the bar. Inclined section Normal section P P Because the stresses are the same throughout the entire bar, the stresses on the sections are uniformly distributed. Inclined Normal P section P section 2 y N θ P x P V The force P can be resolved into components: Normal force N perpendicular to the inclined plane, N = P cos θ Shear force V tangential to the inclined plane V = P sin θ If we know the areas on which the forces act, we can calculate the associated stresses. y y σθ τθ area x area x A A area (A / cos θ) area (A / cos θ) 3 y σθ τθ θ x Force N P cos θ P σ = = = = cos2θ θ Area Area A/cos θ A σ σ = σ cos2θ = x ()1+ cos 2θ θ x 2 σmax = σx occurs when θ = 0° Force −V − P sin θ P τ = = = = − sin θ cos θ θ Area Area A/cos θ A σ τ = −σ sin θ cos θ = − x ()sin 2θ θ x 2 τmax = ± σx/2 occurs when θ = -/+ 45° 4 Introduction to stress elements Stress elements are a useful way to represent stresses acting at some point on a body. Isolate a small element and show stresses acting on all faces. Dimensions are “infinitesimal”, but are drawn to a large scale. y P x σx = P / A Area A y y σx σx = PA/ σx σx = PA/ x x z 5 Maximum stresses on a bar in tension P P a σ x σx = σmax = P / A a Maximum normal stress, Zero shear stress 6 Maximum stresses on a bar in tension P P ab σx/2 θ = 45° τmax = σx/2 σx/2 b Maximum shear stress, Non-zero normal stress 7 Stress Elements and Plane Stress When working with stress elements, keep in mind that only one intrinsic state of stress exists at a point in a stressed body, regardless of the orientation of the element used to portray the state of stress. We are really just rotating axes to represent stresses in a new coordinate system. y y1 x1 θ σx σx x 8 y σy τ Normal stresses σx, σy, σz yx (tension is positive) τyz τxy τzy σx σx x τzx τxz σz Shear stresses τxy = τyx, z τxz = τzx, τyz = τzy σy Sign convention for τab Subscript a indicates the “face” on which the stress acts (positive x “face” is perpendicular to the positive x direction) Subscript b indicates the direction in which the stress acts Strictly σx = σxx, σy = σyy, σz = σzz 9 When an element is in plane stress in the xy plane, only the x and y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0). Such an element could be located on the free surface of a body (no stresses acting on the free surface). Plane stress element in 2D σy σx, σy τyx τ = τ τ xy yx y xy σ = 0 z x σx σx τxy τyx σy 10 Stresses on Inclined Sections y y1 σy τ τyx σy1 y11x τx11y x1 τxy y σx1 x θ σx σx x τ σx1 xy τy11x τyx τx11y σy1 The stress system is known in terms of coordinate system xy. We want to find the stresses in terms of the rotated coordinate system x1y1. Why? A material may yield or fail at the maximum value of σ or τ. This value may occur at some angle other than θ = 0. (Remember that for uni- axial tension the maximum shear stress occurred when θ = 45 degrees. ) 11 Transformation Equations Stresses Forces y y y 1 y1 x1 x τ τ A sec θ 1 θ x11y θ xy11 σx σ A σx1 θ x θ σ A sec θ x x1 x τxy τxy A A tan τyx τθyx σ y σy A tan θ Forces can be found from stresses if Left face has area A. the area on which the stresses act is known. Force components can then Bottom face has area A tan θ. be summed. Inclined face has area A sec θ. 12 y y1 x τ A sec θ 1 θ xy11 σ A x θ σ A sec θ x1 x τxy A τθyx A tan σy A tan θ Sum forces in the x1 direction : σ x1 Asecθ − ()σ x A cosθ − ()τ xy A sinθ − (σ y Atanθ )sinθ − (τ yx Atanθ )cosθ = 0 Sum forces in the y1 direction : τ x1y1 Asecθ + ()σ x A sinθ − ()τ xy A cosθ − (σ y Atanθ )cosθ − (τ yx Atanθ )sinθ = 0 Using τ xy =τ yx and simplifying gives : 2 2 σ x1 = σ x cos θ +σ y sin θ + 2τ xy sinθ cosθ 2 2 τ x1y1 = −()σ x −σ y sinθ cosθ +τ xy ()cos θ − sin θ 13 Using the following trigonometric identities 1+ cos 2θ 1− cos 2θ sin 2θ cos2 θ = sin 2 θ = sinθ cosθ = 2 2 2 gives the transformation equations for plane stress : σ +σ σ −σ σ = x y + x y cos 2θ + τ sin 2θ x1 2 2 xy ()σ −σ HLT, page 108 τ = − x y sin 2θ + τ cos 2θ x1y1 2 xy ° For stresses on the y1 face, substitute θ + 90 for θ : σ +σ σ −σ σ = x y − x y cos 2θ − τ sin 2θ y1 2 2 xy Summing the expressions for x1 and y1 gives : σ x1 +σ y1 = σ x +σ y Can be used to find σy1, instead of eqn above. 14 Example: The state of plane stress at a point is represented by the stress element below. Determine the stresses acting on an element oriented 30° clockwise with respect to the original element. 50 MPa Define the stresses in terms of the established sign convention: y σx = -80 MPa σy = 50 MPa 80 MPa x 80 MPa τxy = -25 MPa We need to find σ , σ , and 25 MPa x1 y1 τ when θ = -30°. 50 MPa x1y1 Substitute numerical values into the transformation equations: σ +σ σ −σ σ = x y + x y cos 2θ + τ sin 2θ x1 2 2 xy − 80 + 50 − 80 − 50 σ = + cos 2()()()− 30° + − 25 sin 2 − 30° = −25.9 MPa x1 2 2 15 σ +σ σ −σ σ = x y − x y cos 2θ − τ sin 2θ y1 2 2 xy −80 + 50 −80 − 50 σ = − cos 2()()()− 30° − − 25 sin 2 − 30° = −4.15 MPa y1 2 2 (σ −σ ) τ = − x y sin 2θ + τ cos 2θ x1y1 2 xy ()− 80 − 50 τ = − sin 2()()()− 30° + − 25 cos 2 − 30° = −68.8 MPa x1y1 2 y y1 4.15 MPa 25.8 MPa Note that σy1 could also be obtained (a) by substituting +60° into the equation for σ or (b) by using the +60° x1 x equation σ + σ = σ + σ -30o x y x1 y1 25.8 MPa 4.15 MPa x1 68.8 MPa (from Hibbeler, Ex. 15.2) 16 Principal Stresses The maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations. σ +σ σ −σ σ = x y + x y cos 2θ + τ sin 2θ x1 2 2 xy dσ σ −σ x1 = x y ()()− 2sin 2θ + τ 2cos 2θ = 0 dθ 2 xy dσ x1 = − ()σ −σ sin 2θ + 2τ cos 2θ = 0 dθ x y xy 2τ xy θ are principal angles associated with tan 2θ p = p σ x −σ y the principal stresses (HLT, page 108) There are two values of 2θp in the range 0-360°, with values differing by 180°. There are two values of θp in the range 0-180°, with values differing by 90°. So, the planes on which the principal stresses act are mutually perpendicular. 17 We can now solve for the principal stresses by substituting for θp in the stress transformation equation for σx1. This tells us which principal stress is associated with which principal angle. 2τ xy tan 2θ p = σ x −σ y σ +σ σ −σ σ = x y + x y cos 2θ + τ sin 2θ x1 2 2 xy 2 ⎛σ −σ ⎞ 2 ⎜ x y ⎟ 2 R = ⎜ ⎟ +τ xy R ⎝ 2 ⎠ τ σ −σ xy cos 2θ = x y p 2R 2θp τ xy sin 2θ p = (σx – σy) / 2 R σ +σ σ −σ ⎛σ −σ ⎞ ⎛τ ⎞ x y x y ⎜ x y ⎟ ⎜ xy ⎟ σ1 = + ⎜ ⎟ + τ xy ⎜ ⎟ 2 2 ⎝ 2R ⎠ ⎝ R ⎠ 18 Substituting for R and re-arranging gives the larger of the two principal stresses: 2 σ +σ ⎛σ −σ ⎞ x y ⎜ x y ⎟ 2 σ1 = + ⎜ ⎟ +τ xy 2 ⎝ 2 ⎠ To find the smaller principal stress, use σ1 + σ2 = σx + σy. 2 σ +σ ⎛σ −σ ⎞ x y ⎜ x y ⎟ 2 σ 2 = σ x +σ y −σ1 = − ⎜ ⎟ +τ xy 2 ⎝ 2 ⎠ These equations can be combined to give: 2 σ +σ ⎛σ −σ ⎞ Principal stresses x y ⎜ x y ⎟ 2 σ1,2 = ± ⎜ ⎟ +τ xy 2 ⎝ 2 ⎠ (HLT page 108) To find out which principal stress goes with which principal angle, we could use the equations for sin θp and cos θp or for σx1.