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The following will be discussed in this chapter: Introduction, transformation of plane stress, principal stresses, maximum shearing stress, mohr’s circle for plane stress, general state of stress, application of mohr’s circle to the three- dimensional analysis of stress, yield criteria for ductile materials under plane stress, fracture criteria for brittle materials under plane stress, stresses in thin-walled pressure vessels.
Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P Beer E Russell Johnston, Jr John T DeWolf Transformations of Stress and Strain Lecture Notes: J Walt Oler Texas Tech University © 2002 The McGraw-Hill Companies, Inc All rights reserved Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Transformations of Stress and Strain Introduction Transformation of Plane Stress Principal Stresses Maximum Shearing Stress Example 7.01 Sample Problem 7.1 Mohr’s Circle for Plane Stress Example 7.02 Sample Problem 7.2 General State of Stress Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress Yield Criteria for Ductile Materials Under Plane Stress Fracture Criteria for Brittle Materials Under Plane Stress Stresses in Thin-Walled Pressure Vessels © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-2 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Introduction • The most general state of stress at a point may be represented by components, σ x ,σ y ,σ z normal stresses τ xy , τ yz , τ zx shearing stresses (Note : τ xy = τ yx , τ yz = τ zy , τ zx = τ xz ) • Same state of stress is represented by a different set of components if axes are rotated • The first part of the chapter is concerned with how the components of stress are transformed under a rotation of the coordinate axes The second part of the chapter is devoted to a similar analysis of the transformation of the components of strain © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-3 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Introduction • Plane Stress - state of stress in which two faces of the cubic element are free of stress For the illustrated example, the state of stress is defined by σ x , σ y , τ xy and σ z = τ zx = τ zy = • State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate • State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at any point of the surface not subjected to an external force © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-4 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Transformation of Plane Stress • Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the x, y, and x’ axes ∑ Fx′ = = σ x′∆A − σ x (∆A cosθ ) cosθ − τ xy (∆A cosθ )sin θ − σ y (∆A sin θ )sin θ − τ xy (∆A sin θ ) cosθ ∑ Fy ′ = = τ x′y ′∆A + σ x (∆A cosθ )sin θ − τ xy (∆A cosθ ) cosθ − σ y (∆A sin θ ) cosθ + τ xy (∆A sin θ )sin θ • The equations may be rewritten to yield σ σ τ σ σ σ σ σ σ σ σ © 2002 The McGraw-Hill Companies, Inc All rights reserved σ 2 σ cos 2θ τ sin 2θ cos 2θ τ sin 2θ sin 2θ τ cos 2θ 7-5 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principal Stresses • The previous equations are combined to yield parametric equations for a circle, (σ x′ − σ ave )2 + τ x2′y′ = R where σ ave = σ x +σ y ⎛σ x −σ y ⎞ ⎟⎟ + τ xy R = ⎜⎜ ⎝ ⎠ • Principal stresses occur on the principal planes of stress with zero shearing stresses σ max,min = tan 2θ p = σ x +σ y 2 ⎛σ x −σ y ⎞ ⎟⎟ + τ xy ± ⎜⎜ ⎝ ⎠ 2τ xy σ x −σ y Note : defines two angles separated by 90o © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-6 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Maximum Shearing Stress Maximum shearing stress occurs for σ x′ = σ ave ⎛σ x −σ y ⎞ ⎟⎟ + τ xy τ max = R = ⎜⎜ ⎝ ⎠ σ x −σ y tan 2θ s = − 2τ xy Note : defines two angles separated by 90o and offset from θ p by 45o σ ′ = σ ave = © 2002 The McGraw-Hill Companies, Inc All rights reserved σ x +σ y 7-7 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 7.01 SOLUTION: • Find the element orientation for the principal stresses from 2τ xy tan 2θ p = σ x −σ y • Determine the principal stresses from σ max,min = σx +σ y ⎛σ x − σ y ⎞ ⎟⎟ + τ xy ± ⎜⎜ ⎝ ⎠ For the state of plane stress shown, determine (a) the principal panes, • Calculate the maximum shearing stress with (b) the principal stresses, (c) the σ σ − ⎛ ⎞ x y maximum shearing stress and the ⎟⎟ + τ xy τ max = ⎜⎜ corresponding normal stress ⎝ ⎠ σx +σ y ′ σ = © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-8 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 7.01 SOLUTION: • Find the element orientation for the principal stresses from 2τ xy 2(+ 40 ) = 1.333 = tan 2θ p = σ x − σ y 50 − (− 10 ) 2θ p = 53.1°, 233.1° σ x = +50 MPa σ x = −10 MPa θ p = 26.6°, 116.6° τ xy = +40 MPa • Determine the principal stresses from σ max,min = σx +σ y = 20 ± ⎛σ x − σ y ⎞ ⎟⎟ + τ xy ± ⎜⎜ ⎠ ⎝ (30)2 + (40)2 σ max = 70 MPa σ = −30 MPa © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-9 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 7.01 • Calculate the maximum shearing stress with ⎛σ x −σ y ⎞ ⎟⎟ + τ xy τ max = ⎜⎜ ⎠ ⎝ = (30)2 + (40)2 τ max = 50 MPa σ x = +50 MPa σ x = −10 MPa τ xy = +40 MPa θ s = θ p − 45 θ s = −18.4°, 71.6° • The corresponding normal stress is σ x + σ y 50 − 10 σ ′ = σ ave = = 2 σ ′ = 20 MPa © 2002 The McGraw-Hill Companies, Inc All rights reserved - 10 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress • If A and B are on the same side of the origin (i.e., have the same sign), then a) the circle defining σmax, σmin, and τmax for the element is not the circle corresponding to transformations within the plane of stress b) maximum shearing stress for the element is equal to half of the maximum stress c) planes of maximum shearing stress are at 45 degrees to the plane of stress © 2002 The McGraw-Hill Companies, Inc All rights reserved - 26 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Yield Criteria for Ductile Materials Under Plane Stress • Failure of a machine component subjected to uniaxial stress is directly predicted from an equivalent tensile test • Failure of a machine component subjected to plane stress cannot be directly predicted from the uniaxial state of stress in a tensile test specimen • It is convenient to determine the principal stresses and to base the failure criteria on the corresponding biaxial stress state • Failure criteria are based on the mechanism of failure Allows comparison of the failure conditions for a uniaxial stress test and biaxial component loading © 2002 The McGraw-Hill Companies, Inc All rights reserved - 27 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Yield Criteria for Ductile Materials Under Plane Stress Maximum shearing stress criteria: Structural component is safe as long as the maximum shearing stress is less than the maximum shearing stress in a tensile test specimen at yield, i.e., σ τ max < τ Y = Y For σa and σb with the same sign, τ max = σa or σb σ < Y 2 For σa and σb with opposite signs, τ max = © 2002 The McGraw-Hill Companies, Inc All rights reserved σa −σb σ < Y - 28 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Yield Criteria for Ductile Materials Under Plane Stress Maximum distortion energy criteria: Structural component is safe as long as the distortion energy per unit volume is less than that occurring in a tensile test specimen at yield ud < uY 2 σ a − σ aσ b + σ b2 < σ Y − σ Y × + 02 6G 6G ( ) ( ) σ a2 − σ aσ b + σ b2 < σ Y2 © 2002 The McGraw-Hill Companies, Inc All rights reserved - 29 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Fracture Criteria for Brittle Materials Under Plane Stress Brittle materials fail suddenly through rupture or fracture in a tensile test The failure condition is characterized by the ultimate strength σU Maximum normal stress criteria: Structural component is safe as long as the maximum normal stress is less than the ultimate strength of a tensile test specimen σ a < σU σ b < σU © 2002 The McGraw-Hill Companies, Inc All rights reserved - 30 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses in Thin-Walled Pressure Vessels • Cylindrical vessel with principal stresses σ1 = hoop stress σ2 = longitudinal stress • Hoop stress: ∑ Fz = = σ 1(2t ∆x ) − p (2r ∆x ) σ1 = pr t • Longitudinal stress: ( ) ∑ Fx = = σ (2π rt ) − p π r pr σ2 = 2t σ = 2σ © 2002 The McGraw-Hill Companies, Inc All rights reserved - 31 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses in Thin-Walled Pressure Vessels • Points A and B correspond to hoop stress, σ1, and longitudinal stress, σ2 • Maximum in-plane shearing stress: τ max(in − plane) = σ = pr 4t • Maximum out-of-plane shearing stress corresponds to a 45o rotation of the plane stress element around a longitudinal axis τ max = σ = © 2002 The McGraw-Hill Companies, Inc All rights reserved pr 2t - 32 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses in Thin-Walled Pressure Vessels • Spherical pressure vessel: σ1 = σ = pr 2t • Mohr’s circle for in-plane transformations reduces to a point σ = σ = σ = constant τ max(in -plane) = • Maximum out-of-plane shearing stress τ max = 12 σ = © 2002 The McGraw-Hill Companies, Inc All rights reserved pr 4t - 33 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Transformation of Plane Strain • Plane strain - deformations of the material take place in parallel planes and are the same in each of those planes • Plane strain occurs in a plate subjected along its edges to a uniformly distributed load and restrained from expanding or contracting laterally by smooth, rigid and fixed supports components of strain : ε x ε y γ xy (ε z = γ zx = γ zy = 0) • Example: Consider a long bar subjected to uniformly distributed transverse loads State of plane stress exists in any transverse section not located too close to the ends of the bar © 2002 The McGraw-Hill Companies, Inc All rights reserved - 34 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Transformation of Plane Strain • State of strain at the point Q results in different strain components with respect to the xy and x’y’ reference frames ε (θ ) = ε x cos θ + ε y sin θ + γ xy sin θ cosθ ε OB = ε (45°) = 12 (ε x + ε y + γ xy ) γ xy = 2ε OB − (ε x + ε y ) • Applying the trigonometric relations used for the transformation of stress, εx + ε y εx − ε y γ xy ε x′ = + cos 2θ + sin 2θ ε y′ = γ x′y′ © 2002 The McGraw-Hill Companies, Inc All rights reserved 2 εx + ε y εx − ε y γ xy =− − εx − ε y 2 cos 2θ − sin 2θ + γ xy 2 sin 2θ cos 2θ - 35 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Mohr’s Circle for Plane Strain • The equations for the transformation of plane strain are of the same form as the equations for the transformation of plane stress - Mohr’s circle techniques apply • Abscissa for the center C and radius R , ε ave = εx + ε y ⎛ ε x − ε y ⎞ ⎛ γ xy ⎞ ⎟⎟ ⎟⎟ + ⎜⎜ R = ⎜⎜ ⎝ ⎠ ⎝ ⎠ 2 • Principal axes of strain and principal strains, γ xy tan 2θ p = εx − ε y ε max = ε ave + R ε = ε ave − R • Maximum in-plane shearing strain, γ max = R = © 2002 The McGraw-Hill Companies, Inc All rights reserved (ε x − ε y )2 + γ xy2 - 36 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Three-Dimensional Analysis of Strain • Previously demonstrated that three principal axes exist such that the perpendicular element faces are free of shearing stresses • By Hooke’s Law, it follows that the shearing strains are zero as well and that the principal planes of stress are also the principal planes of strain • Rotation about the principal axes may be represented by Mohr’s circles © 2002 The McGraw-Hill Companies, Inc All rights reserved - 37 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Three-Dimensional Analysis of Strain • For the case of plane strain where the x and y axes are in the plane of strain, - the z axis is also a principal axis - the corresponding principal normal strain is represented by the point Z = or the origin • If the points A and B lie on opposite sides of the origin, the maximum shearing strain is the maximum in-plane shearing strain, D and E • If the points A and B lie on the same side of the origin, the maximum shearing strain is out of the plane of strain and is represented by the points D’ and E’ © 2002 The McGraw-Hill Companies, Inc All rights reserved - 38 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Three-Dimensional Analysis of Strain • Consider the case of plane stress, σ x = σa σ y = σb σ z = • Corresponding normal strains, σ νσ εa = a − b E εb = − E νσ a σ + b E E ν ν (ε a + ε b ) ε c = − (σ a + σ b ) = − −ν E • Strain perpendicular to the plane of stress is not zero • If B is located between A and C on the Mohr-circle diagram, the maximum shearing strain is equal to the diameter CA © 2002 The McGraw-Hill Companies, Inc All rights reserved - 39 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Measurements of Strain: Strain Rosette • Strain gages indicate normal strain through changes in resistance • With a 45o rosette, εx and εy are measured directly γxy is obtained indirectly with, γ xy = 2ε OB − (ε x + ε y ) • Normal and shearing strains may be obtained from normal strains in any three directions, ε1 = ε x cos θ1 + ε y sin θ1 + γ xy sin θ1 cosθ1 ε = ε x cos θ + ε y sin θ + γ xy sin θ cosθ ε = ε x cos θ + ε y sin θ + γ xy sin θ cosθ © 2002 The McGraw-Hill Companies, Inc All rights reserved - 40 ...Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Transformations of Stress and Strain Introduction Transformation of Plane Stress Principal Stresses Maximum Shearing Stress Example... of the chapter is devoted to a similar analysis of the transformation of the components of strain © 2002 The McGraw-Hill Companies, Inc All rights reserved 7-3 Third Edition MECHANICS OF MATERIALS. .. Introduction • Plane Stress - state of stress in which two faces of the cubic element are free of stress For the illustrated example, the state of stress is defined by σ x , σ y , τ xy and σ z = τ zx