MECHANICS OF MATERIALS

Transcription

1 CHAPTER 7 Transformations MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech Universit of Stress and Strain 006 The McGraw-Hill Companies, Inc. All rights reserved. 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 7.0 Sample Problem 7. General State of Stress Application of Mohr s Circle to the Three-Dimensional Analsis of Stress Yield Criteria for Ductile Materials Under Plane Stress Fracture Criteria for Brittle Materials Under Plane Stress Stresses in Thin-Walled Pressure Vessels 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-1

2 Introduction The most general state of stress at a point ma be represented b 6 components, σ x, σ, σ z τ x, τ z, τ zx normal stresses shearing stresses (Note : τ x = τ x, τ z = τ z, τ zx = τ xz ) Same state of stress is represented b 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 analsis of the transformation of the components of strain. 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-3 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 b σ, σ, τ and σ τ = τ 0. x x z = zx z = 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 an point of the surface not subjected to an external force. 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

3 Transformation of Plane Stress Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the x,, and x axes. Fx = 0 = σ x A σ x( Acosθ ) cosθ τ x( Acosθ ) sinθ σ ( Asinθ ) sinθ τ x( Asinθ ) cosθ F = 0 = τ x A + σ x( Acosθ ) sinθ τ x( Acosθ ) cosθ σ ( Asinθ ) cosθ + τ x( Asinθ ) sinθ The equations ma be rewritten to ield σ x + σ σ σ x = + cos θ + τ x sin θ σ x + σ σ σ = cos θ τ x sin θ σ τ x = sin θ + τ x cosθ 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-5 Principal Stresses The previous equations are combined to ield parametric equations for a circle, ( σ σ ) x ave + τ x = R where σ x + σ σ ave = R = σ + τ x Principal stresses occur on the principal planes of stress with zero shearing stresses. σ x + σ σ max,min = ± τ x tan θ p = σ σ + τ x o Note : defines two angles separated b The McGraw-Hill Companies, Inc. All rights reserved

4 Maximum Shearing Stress Maximum shearing stress occurs for σ τ max = R = x + τ σ tan θ s = τ x o offset fromθ p b 45 σ x + σ σ = σ ave = σ x = σ ave o Note : defines two angles separated b 90 and 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-7 Example 7.01 For the state of plane stress shown, determine (a) the principal planes, (b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress. SOLUTION: Find the element orientation for the principal stresses from τ x tan θ p = σ x σ Determine the principal stresses from σ x + σ σ σ max,min = ± + τ x Calculate the maximum shearing stress with σ τ max = + τ x σ x + σ σ = 006 The McGraw-Hill Companies, Inc. All rights reserved

5 Example 7.01 σ x = + 50MPa σ x = 10MPa τ x = + 40MPa SOLUTION: Find the element orientation for the principal stresses from τ x ( + 40) tan θ p = = = σ x σ 50 ( 10) θ = 53.1, 33.1 p θ p = 6.6, Determine the principal stresses from σ σ x + σ σ max,min = ± + τ x = 0 ± σ max = 70MPa σ min = 30MPa ( 30) + ( 40) 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-9 Example 7.01 σ x = + 50MPa σ x = 10MPa τ x = + 40MPa Calculate the maximum shearing stress with τ σ max = + τ x = τ max = 50MPa θ s = θ p ( 30) + ( 40) 45 θ s = 18.4, The corresponding normal stress is σ x + σ σ = σ ave = = σ = 0MPa 006 The McGraw-Hill Companies, Inc. All rights reserved

6 Sample Problem 7.1 A single horizontal force P of 150 lb magnitude is applied to end D of lever ABD. Determine (a) the normal and shearing stresses on an element at point H having sides parallel to the x and axes, (b) the principal planes and principal stresses at the point H. SOLUTION: Determine an equivalent force-couple sstem at the center of the transverse section passing through H. Evaluate the normal and shearing stresses at H. Determine the principal planes and calculate the principal stresses. 006 The McGraw-Hill Companies, Inc. All rights reserved Sample Problem 7.1 SOLUTION: Determine an equivalent force-couple sstem at the center of the transverse section passing through H. P = 150lb T = ( 150lb)( 18in) =.7 kip in = ( 150lb)( 10in) = 1.5kip in M x Evaluate the normal and shearing stresses at H. Mc ( 1.5kip in)( 0.6in) σ = + = + I 1 4 π ( 0.6in) 4 Tc (.7 kip in)( 0.6in) τ x = + = + J 1π ( 0.6in) 4 σ x = 0 σ = ksi τ = ksi 006 The McGraw-Hill Companies, Inc. All rights reserved

7 Sample Problem 7.1 Determine the principal planes and calculate the principal stresses. τ x ( 7.96) tan θ p = = = 1.8 σ x σ θ = 61.0,119 σ p θ p = 30.5, σ x + σ σ x σ max,min = ± + τ x = ± σ max = ksi σ min = 4.68ksi ( 7.96) 006 The McGraw-Hill Companies, Inc. All rights reserved Mohr s Circle for Plane Stress With the phsical significance of Mohr s circle for plane stress established, it ma be applied with simple geometric considerations. Critical values are estimated graphicall or calculated. For a known state of plane stress σ x, σ, τ plot the points X and Y and construct the circle centered at C. σ x + σ σ σ ave = R = + τ x The principal stresses are obtained at A and B. σ = σ ± R max,min ave τ x tan θ p = σ The direction of rotation of Ox to Oa is the same as CX to CA. x 006 The McGraw-Hill Companies, Inc. All rights reserved

8 Mohr s Circle for Plane Stress With Mohr s circle uniquel defined, the state of stress at other axes orientations ma be depicted. For the state of stress at an angle θ with respect to the x axes, construct a new diameter X Y at an angle θ with respect to XY. Normal and shear stresses are obtained from the coordinates X Y. 006 The McGraw-Hill Companies, Inc. All rights reserved Mohr s Circle for Plane Stress Mohr s circle for centric axial loading: P σ x =, σ = τ x = 0 A P σ x = σ = τ x = A Mohr s circle for torsional loading: Tc Tc σ x = σ = 0 τ x = σ x = σ = τ x = 0 J J 006 The McGraw-Hill Companies, Inc. All rights reserved

9 Example 7.0 For the state of plane stress shown, (a) construct Mohr s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding normal stress. 006 The McGraw-Hill Companies, Inc. All rights reserved. SOLUTION: Construction of Mohr s circle σ x + σ ( 50) + ( 10) σ ave = = = 0MPa CF = 50 0 = 30MPa FX = 40MPa R = CX = ( 30) + ( 40) = 50MPa 7-17 Example 7.0 Principal planes and stresses σ max = OA = OC + CA = σ max = 70MPa σ min = OB = OC BC = 0 50 σ min = 30MPa FX = CP θ p = 53.1 tan θ p = θ p = The McGraw-Hill Companies, Inc. All rights reserved

10 Example 7.0 Maximum shear stress θ s = θ p + 45 θ s = τ max = R τ max = 50 MPa σ = σ ave σ = 0 MPa 006 The McGraw-Hill Companies, Inc. All rights reserved Sample Problem 7. For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained b rotating the given element counterclockwise through 30 degrees. SOLUTION: Construct Mohr s circle σ x + σ σ ave = = = 80MPa R = ( CF ) + ( FX ) = ( 0) + ( 48) = 5MPa 006 The McGraw-Hill Companies, Inc. All rights reserved

11 Sample Problem 7. Principal planes and stresses XF 48 tan θ p = = =.4 CF 0 θ p = 67.4 θ p = 33.7 clockwise σ max = OA = OC + CA = σ max = +13MPa σ max = OA = OC BC = 80 5 σ min = +8 MPa 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-1 Sample Problem 7. Stress components after rotation b 30 o Points X and Y on Mohr s circle that correspond to stress components on the rotated element are obtained b rotating XY counterclockwise through θ = The McGraw-Hill Companies, Inc. All rights reserved. φ = = 5.6 σ x = OK = OC KC = 80 5cos5.6 σ = OL = OC + CL = cos5.6 τ x = KX = 5sin 5.6 σ x = MPa σ = MPa τ x = 41.3MPa 7-11

12 General State of Stress Consider the general 3D state of stress at a point and the transformation of stress from element rotation State of stress at Q defined b: σ, σ, σ, τ, τ, τ Consider tetrahedron with face perpendicular to the line QN with direction cosines: λ, λ, λ The requirement F n = 0 leads to, σ n = σ xλx + σ λ + σ zλz + τ xλxλ + τ zλλz + τ zxλzλx Form of equation guarantees that an element orientation can be found such that n = aλa + bλb σ σ σ + σ cλc These are the principal axes and principal planes and the normal stresses are the principal stresses. x x z z x z zx 006 The McGraw-Hill Companies, Inc. All rights reserved. 7-3 Application of Mohr s Circle to the Three- Dimensional Analsis of Stress Transformation of stress for an element rotated around a principal axis ma be represented b Mohr s circle. Points A, B, and C represent the principal stresses on the principal planes (shearing stress is zero) 006 The McGraw-Hill Companies, Inc. All rights reserved. The three circles represent the normal and shearing stresses for rotation around each principal axis. Radius of the largest circle ields the maximum shearing stress. τ 1 max = σ max σ min 7-4 1

13 Application of Mohr s Circle to the Three- Dimensional Analsis of Stress In the case of plane stress, the axis perpendicular to the plane of stress is a principal axis (shearing stress equal zero). If the points A and B (representing the principal planes) are on opposite sides of the origin, then a) the corresponding principal stresses are the maximum and minimum normal stresses for the element b) the maximum shearing stress for the element is equal to the maximum inplane shearing stress c) planes of maximum shearing stress are at 45 o to the principal planes. 006 The McGraw-Hill Companies, Inc. All rights reserved